Functions – JEE Main MathematicsPractice Questions & PYQs

The domain is determined by the intersection of constraints from the inverse sine function and the square root denominator. The argument of $ext{sin}^{-1}(x-3)$ must lie within $[-1, 1]$, resulting in $2 ext{leq} x ext{leq} 4$, while the denominator $9-x^2 > 0$ requires $-3 < x < 3$. Combining these two requirements, the valid values for $x$ are restricted to the interval $[2, 3)$.

For the function to be defined, the constraints require $7-x \ge x-3 \ge 0$, which simplifies to $3 \le x \le 5$ for integer values of $x$. Evaluating the expression at $x=3$, $x=4$, and $x=5$ yields the output values for the range. This process results in a set of values equal to $\{1, 2, 3\}$.

The function $f(x) = ext{sin } x - ext{sqrt}{3} ext{cos } x + 1$ can be expressed in the form $r ext{sin}(x - heta) + c$, where $r = ext{sqrt}{1^2 + (- ext{sqrt}{3})^2} = 2$. This simplifies the expression to 2 ext{sin}(x - rac{rac{\pi}{3}}) + 1, which oscillates between $2(-1)+1 = -1$ and $2(1)+1 = 3$. Therefore, for the function to be onto, the set $S$ must equal the interval $[-1, 3]$.

A function is classified as odd if it satisfies the condition $f(-x) = -f(x)$. Upon substitution, $f(-x) = \log(-x + \sqrt{(-x)^2 + 1})$, which simplifies to $\log(\sqrt{x^2 + 1} - x)$. By rationalizing the argument as $\frac{(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)}{\sqrt{x^2 + 1} + x}$, the expression becomes $\log\left(\frac{1}{x + \sqrt{x^2 + 1}}\right)$, equivalent to $-\log(x + \sqrt{x^2 + 1})$, confirming the function is odd.

The domain is determined by ensuring the denominator $4-x^2$ is non-zero and the logarithmic argument $x^{3-}x$ is strictly positive. Solving $x^{3-}x > 0$ factors to $x(x-1)(x+1) > 0$, which holds for $x \in (-1, 0) \cup (1, \infty )$. Applying the condition $x \neq \pm 2$ restricts the valid region by excluding the point $2$, resulting in the interval $(-1, 0) \cup (1, 2) \cup (2, \infty )$.

The linear functional equation $f(x+y) = f(x) + f(y)$ implies a function of the form $f(x) = cx$, and applying the condition $f(1) = 7$ identifies the constant as $7$, making $f(x) = 7x$. The summation $\sum_{r=1}^{n} f(r)$ becomes $7 \sum_{r=1}^{n} r$. Utilizing the standard formula for the sum of the first $n$ integers, this resolves to $\frac{7 n(n+1)}{2}$.

A function is one-one if every distinct input yields a unique output, and onto if every integer in the codomain has a corresponding natural number in the domain. In this case, odd natural numbers map to $\frac{n-1}{2}$, generating all non-negative integers, while even natural numbers map to $-\frac{n}{2}$, covering all negative integers. Because these two branches produce distinct, unique integer outputs that span the set of all integers, the function is both one-one and onto.

The period of a trigonometric function raised to a power depends on the parity of the exponent, where even exponents cause the function to repeat every $\pi$ interval. Specifically, since $\sin^2(\theta + \pi) = (\sin(\theta + \pi))^2 = (-\sin \theta)^2 = \sin^2 \theta$, the cycle repeats itself after every $\pi$ radians.

The definition of the inverse sine function requires its argument to lie within the closed interval $[-1, 1]$. Setting the inequality $-1 \le \log_3(\frac{x}{3}) \le 1$ and applying exponentiation base 3 results in $3^{-1} \le \frac{x}{3} \le 3^1$, which simplifies to $\frac{1}{3} \le \frac{x}{3} \le 3$. Multiplying the entire inequality by 3 yields $1 \le x \le 9$, establishing the domain as $[1, 9]$.

A function is periodic if there exists a fixed constant $T$ such that $f(x+T) = f(x)$ for all $x$. While trigonometric components like $\cos^2 x$ are periodic, the term $\cos\sqrt{x}$ contains a square root that causes the function values to change inconsistently, preventing regular repetition. Since the sum of a periodic and a non-periodic function is non-periodic, $\cos\sqrt{x} + \cos^2 x$ does not satisfy the requirements for periodicity.

For any non-integer $x$, the ceiling function expression simplifies as $\lceil x \rceil - x = 1 - \{x\}$, where $\{x\}$ represents the fractional part of $x$. This transforms the function into $f(x) = \frac{1}{\sqrt{1-\{x\}}}$, which is defined for all $x \notin \mathbb{Z}$, establishing the domain $A = \mathbb{R} - \mathbb{Z}$. Since $0 \leq \{x\} < 1$, the function outputs values in the range $B = (1, \infty )$, and comparing these sets confirms that $A \cap B = (1, \infty ) - N$ is true.

Analyzing the piecewise definition for the domain $N$ reveals how inputs map to unique outputs. Inputs $n o 1, 5, 9, 13, ext{...}$ map to all odd numbers, inputs $n o 3, 7, 11, 15, ext{...}$ map to even numbers that are not multiples of 4, and inputs $n o 2, 4, 6, 8, ext{...}$ map to all multiples of 4. Because every natural number is generated exactly once and each input produces a distinct output, the function is both one-to-one and onto.

Performing the column operation $C_1 \rightarrow C_1+C_2+C_3$ simplifies the determinant by allowing the factor $(2+\sin 2x)$ to be pulled out, leaving a determinant that evaluates to $1$. For $x \in [\pi/6, \pi/3]$, the expression $2+\sin 2x$ varies from $2+\sqrt{3}/2$ to $3$, giving a minimum value $\beta = 2+\sqrt{3}/2$ and a maximum $\alpha=3$. Calculating the required expression yields $\beta^{2-}2\sqrt{\alpha} = (2+\frac{\sqrt{3}}{2})^2 - 2\sqrt{3} = \frac{19+8\sqrt{3}}{4} - \frac{8\sqrt{3}}{4} = \frac{19}{4}$.